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User blog:Holomanga/electromagnetic waves
it's 3am and i'm just about ready to dielectric just placing this down here so i remember to make a blog post on it it's going to be great did you know there's a thing called a Poynting vector sounds like a kabbalistic pun any way yeah EM waves time later constantly refresh your browser Phasors So imagine you have something that varies sinusoidally. You can write this as X = \mathrm{Re} \left\{ X_0 e^{i\omega t} \right\} . Now say you want to phase shift it by an angle φ, then it becomes Y = \mathrm{Re} \left\{ X_0 e^{i\omega t} e^{i\phi} \right\} . But wait! e^{i \phi} is just a constant, so you can roll it into the other constant, and just get Y = \mathrm{Re} \left\{ Y_0 e^{i\omega t} \right\} . Note that in general the constants are complex. Now, if the operations you're using are all linear (which they will be because I will shout at you if you do nonlinear things), then you can just pull out the real part operator and then put it in later. When you include both the real and imaginary parts of your sinusoidally varying thing, you get a complex quantity called a phasor. \widetilde{Y} = \widetilde{Y}_0 e^{i \omega t} Here, the tilde reminds you that it's constant - you need to take the real part to get the actual physical things. The imaginary part represents a phase difference of sorts. (The best part about phasors is that the real part is actually real, and the imaginary part is actually imaginary) Electromagnetic Waves in a Vacuum Recall Maxwell's equations in a vacuum, where there is no free charge and there sure as hell isn't any free current : \nabla \cdot \vec{\widetilde{E}} = 0 : \nabla \cdot \vec{\widetilde{B}} = 0 : \nabla \times \vec{\widetilde{E}} = - \frac{\partial \vec{\widetilde{B}}}{\partial t} : \nabla \times \vec{\widetilde{B}} = \mu_0 \varepsilon_0 \frac{\partial \vec{\widetilde{E}}}{\partial t} Here, all of the E and B fields are being written as phasor fields. I'm kinda assuming that they'll end up as waves but because they are all homogeneous differential equations yeah they'll definitely be waves I'm pretty sure that the results are exponentials or something that's gotta be a theorem. Anyway let's, just for fun, take the curl of the curl of the electric field two lower expressions : \nabla \times \nabla \times \vec{\widetilde{E}} = - \frac{\partial}{\partial t} \nabla \times \vec{\widetilde{B}} We remember from vector calculus that the curl of the curl is the gradient of the divergence minus the laplacian, and we also know what the curl of B is, so : \nabla \nabla \cdot \vec{\widetilde{E}} - \nabla^2 \vec{\widetilde{E}} = - \mu_0 \varepsilon_0 \frac{\partial^2 \vec{\widetilde{E}}}{\partial t^2} But hey the divergence of E is zero right so we just got : \nabla^2 \vec{\widetilde{E}} = \mu_0 \varepsilon_0 \frac{\partial^2 \vec{\widetilde{E}}}{\partial t^2} Now this looks just like the wave equation so let's try to solve this with some wave solution \vec{\widetilde{E}} = \vec{\widetilde{E}}_0 e^{i\left(kz - \omega t\right)} with angular wavenumber k and angular frequency ω. Also we're reorienting the axes so the electric field is only propagating in the z direction because taking the derivatives of the exponential of the dot product of vectors is kinda eh and I think there's a theorem you need to prove before you're legitimately allowed to do it. Rotating your whole system is easier. : k^2 \vec{\widetilde{E}}_0 e^{i\left(kz - \omega t\right)} = \omega^2 \mu_0 \varepsilon_0 \vec{\widetilde{E}}_0 e^{i\left(kz - \omega t\right)} : \frac{\omega^2}{k^2}= \frac{1}{\mu_0 \varepsilon_0} Now wait, the left hand side is the square of the velocity of the wave! So the speed of an electromagnetic wave is the same, no matter what your wavenumber and frequency is - that's what Maxwell's equations say. In fact, it looks the same to you no matter where you are or how fast you're going. This is how special relativity is derived. And, when you plug in the measured values for the electric and magnetic constants, you get the speed of light. Light moves at the speed of light! How novel! (Back in the day, people didn't know that light was EM waves, but it turns out that it totally is.) Now let's do the same with the B field and whoops my hand slipped and I did it accidentally \vec{\widetilde{B}} = \frac{1}{c} \left( \hat{z} \times \vec{\widetilde{E}}_0 \right) e^{i\left(kz - \omega t\right)} The magnetic field is the electric field, divided by the speed of light (for an EM wave), pointing in a different direction. Check out those dimensions! in an ideal dielectric Maxwell's equations in a dielectric are super easy to remember. : \nabla \cdot \vec{\widetilde{E}} = 0 : \nabla \cdot \vec{\widetilde{B}} = 0 : \nabla \times \vec{\widetilde{E}} = - \frac{\partial \vec{\widetilde{B}}}{\partial t} : \nabla \times \vec{\widetilde{B}} = \mu \varepsilon \frac{\partial \vec{\widetilde{E}}}{\partial t} Turns out you just use the permittivity and permeability of the medium that the waves are going through, instead of for a vacuum. This gives \frac{k}{\omega} = \frac{1}{\sqrt{\epsilon \mu}} , as you might expect. bouncing your waves off a dielectric in an ohmic conductor In an ohmic conductor, the current flowing is related to the electric field by \vec J = \sigma \vec E . This is V = IR, but on a microscopic scale: σ is the conductivity, corresponding to 1/R, J is the current density, corresponding to I, and E is the electric field, corresponding to V (with various factors of 1/volume and stuff inserted in; it all checks out in the end). This means our new Maxwell's equations are : \nabla \cdot \vec{\widetilde{E}} = \frac{\rho_f}{\epsilon} : \nabla \cdot \vec{\widetilde{B}} = 0 : \nabla \times \vec{\widetilde{E}} = - \frac{\partial \vec{\widetilde{B}}}{\partial t} : \nabla \times \vec{\widetilde{B}} = \mu \sigma \vec E + \mu \varepsilon \frac{\partial \vec{\widetilde{E}}}{\partial t} We're going to have to do something about that free charge in the conductor. I remember something about how current makes charges flow away. Let's look at the continuity equation for free charge \nabla \cdot \vec{J}_f = - \frac{\partial \rho_f}{\partial t} But hey, we know the current in terms of the electric field because of Ohm's law, so let's try something \sigma \nabla \cdot \vec E = - \frac{\partial \rho_f}{\partial t} But wait \frac{\partial \rho_f}{\partial t} = -\frac{ \sigma}{\varepsilon} \rho_f So \rho_f = \rho_0 e^{-\frac{\sigma}{\varepsilon} t} so after a sufficiently long time we just don't need to think about the free charge. That's a relief! in a plasma Category:Blog posts